(D) Let $P$ be the population at time $t$. Given $\frac{dP}{dt} \propto P$,so $\frac{dP}{dt} = kP$. Integrating,we get $\ln P = kt + C$. At $t = 0$,$P

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(D) Let $P$ be the population at time $t$. Given $\frac{dP}{dt} \propto P$,so $\frac{dP}{dt} = kP$. Integrating,we get $\ln P = kt + C$. At $t = 0$,$P = 20$,so $C = \ln 20$. Thus,$\ln P = kt + \ln 20$. At $t = 30$,$P = 40$,so $\ln 40 = 30k + \ln 20$,which gives $30k = \ln 2$,or $k = \frac{\ln 2}{30}$. We need to find $P$ at $t = 30 + 15 = 45$ years. $\ln P = \left(\frac{\ln 2}{30}\right) \times 45 + \ln 20 = 1.5 \ln 2 + \ln 20 = \ln(2^{1.5} \times 20)$. $P = 20 \times 2^{1.5} = 20 \times 2 \times \sqrt{2} = 40 \times 1.41 = 56.4$ lakhs.

Class 12 Mathematics Differential Equations Application of differential equations

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The rate at which the population of a city increases varies as the population present. Within the period of $30$ years,the population grew from $20$ lakhs to $40$ lakhs. Then,the population after a further period of $15$ years will be (Take $\sqrt{2} = 1.41$) (in $lakhs$)

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A
$56$
B
$60$
C
$57.4$
D
$56.4$

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The rate at which the population of a city increases varies as the population present. Within the period of $30$ years,the population grew from $20$ lakhs to $40$ lakhs. Then,the population after a further period of $15$ years will be (Take $\sqrt{2} = 1.41$) (in $lakhs$)

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In a certain culture of bacteria,the rate of increase is proportional to the number present. If there are $10^4$ at the end of $3$ hours and $4 \cdot 10^4$ at the end of $5$ hours,then there were $\qquad$ in the beginning.

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